3. evaluate the limits at infinity.
a) $ \lim _ { x \
ightarrow \infty } \frac { x ^ { 4 } - 3 x ^ { 3 } + 1 } { x ^ { 3 } - 2 x ^ { 4 } + 2 x } = $
b) $ \lim _ { x \
ightarrow - \infty } \frac { x ^ { 3 } + 7 x - 9 } { x ^ { 2 } - 5 x + 6 } = $
c) $ \lim _ { x \
ightarrow \infty } \frac { ( x ^ { 2 } + 5 x + 1 ) ( x + 2 ) } { x ^ { 4 } - 2 x ^ { 2 } + 2 x } = $
d) $ \lim _ { x \
ightarrow \infty } \frac { 2 x + 3 } { x + \sqrt { 4 x ^ { 2 } + 3 } } = $
$ \lim _ { x \
ightarrow - \infty } \frac { 2 x + 3 } { x + \sqrt { 4 x ^ { 2 } + 3 } } = $

Question

Accepted Answer

# Explanation:
## Step1: Divide numerator and denominator by highest - power of $x$
For part (a):
Divide numerator $x^{4}-3x^{3}+1$ and denominator $x^{3}-2x^{4}+2x$ by $x^{4}$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{x^{4}-3x^{3}+1}{x^{3}-2x^{4}+2x}&=\lim_{x
ightarrow\infty}\frac{\frac{x^{4}}{x^{4}}-\frac{3x^{3}}{x^{4}}+\frac{1}{x^{4}}}{\frac{x^{3}}{x^{4}}-\frac{2x^{4}}{x^{4}}+\frac{2x}{x^{4}}}\
&=\lim_{x
ightarrow\infty}\frac{1 - \frac{3}{x}+\frac{1}{x^{4}}}{\frac{1}{x}-2+\frac{2}{x^{3}}}
\end{align*}
$$
## Step2: Use $\lim_{x
ightarrow\infty}\frac{1}{x^{n}} = 0$ for $n>0$
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x}=0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{3}} = 0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{4}}=0$
$$
\lim_{x
ightarrow\infty}\frac{1 - \frac{3}{x}+\frac{1}{x^{4}}}{\frac{1}{x}-2+\frac{2}{x^{3}}}=\frac{1-0 + 0}{0 - 2+0}=-\frac{1}{2}
$$

## Step3: For part (b)
Divide numerator $x^{3}+7x - 9$ and denominator $x^{2}-5x + 6$ by $x^{2}$
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{x^{3}+7x - 9}{x^{2}-5x + 6}&=\lim_{x
ightarrow-\infty}\frac{\frac{x^{3}}{x^{2}}+\frac{7x}{x^{2}}-\frac{9}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{5x}{x^{2}}+\frac{6}{x^{2}}}\
&=\lim_{x
ightarrow-\infty}\frac{x+\frac{7}{x}-\frac{9}{x^{2}}}{1-\frac{5}{x}+\frac{6}{x^{2}}}
\end{align*}
$$
As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{1}{x}=0$, $\lim_{x
ightarrow-\infty}\frac{1}{x^{2}} = 0$
$$
\lim_{x
ightarrow-\infty}\frac{x+\frac{7}{x}-\frac{9}{x^{2}}}{1-\frac{5}{x}+\frac{6}{x^{2}}}=-\infty
$$

## Step4: For part (c)
First, expand the numerator $(x^{2}+5x + 1)(x + 2)=x^{3}+2x^{2}+5x^{2}+10x+x + 2=x^{3}+7x^{2}+11x + 2$
Divide numerator $x^{3}+7x^{2}+11x + 2$ and denominator $x^{4}-2x^{2}+2x$ by $x^{4}$
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{(x^{2}+5x + 1)(x + 2)}{x^{4}-2x^{2}+2x}&=\lim_{x
ightarrow\infty}\frac{\frac{x^{3}}{x^{4}}+\frac{7x^{2}}{x^{4}}+\frac{11x}{x^{4}}+\frac{2}{x^{4}}}{\frac{x^{4}}{x^{4}}-\frac{2x^{2}}{x^{4}}+\frac{2x}{x^{4}}}\
&=\lim_{x
ightarrow\infty}\frac{\frac{1}{x}+\frac{7}{x^{2}}+\frac{11}{x^{3}}+\frac{2}{x^{4}}}{1-\frac{2}{x^{2}}+\frac{2}{x^{3}}}
\end{align*}
$$
Using $\lim_{x
ightarrow\infty}\frac{1}{x^{n}}=0$ for $n > 0$, we get $\lim_{x
ightarrow\infty}\frac{\frac{1}{x}+\frac{7}{x^{2}}+\frac{11}{x^{3}}+\frac{2}{x^{4}}}{1-\frac{2}{x^{2}}+\frac{2}{x^{3}}}=0$

## Step5: For part (d) (when $x
ightarrow\infty$)
Divide numerator and denominator by $x$ ($x>0$, so $\sqrt{x^{2}}=x$)
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{2x + 3}{x+\sqrt{4x^{2}+3}}&=\lim_{x
ightarrow\infty}\frac{2+\frac{3}{x}}{1+\sqrt{4+\frac{3}{x^{2}}}}\
\end{align*}
$$
Using $\lim_{x
ightarrow\infty}\frac{1}{x}=0$ and $\lim_{x
ightarrow\infty}\frac{1}{x^{2}}=0$
$$
\lim_{x
ightarrow\infty}\frac{2+\frac{3}{x}}{1+\sqrt{4+\frac{3}{x^{2}}}}=\frac{2 + 0}{1+\sqrt{4+0}}=\frac{2}{3}
$$

## Step6: For part (d) (when $x
ightarrow-\infty$)
Divide numerator and denominator by $x$ ($x<0$, so $\sqrt{x^{2}}=-x$)
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{2x + 3}{x+\sqrt{4x^{2}+3}}&=\lim_{x
ightarrow-\infty}\frac{2+\frac{3}{x}}{1-\sqrt{4+\frac{3}{x^{2}}}}\
\end{align*}
$$
Using $\lim_{x
ightarrow-\infty}\frac{1}{x}=0$ and $\lim_{x
ightarrow-\infty}\frac{1}{x^{2}}=0$
$$
\lim_{x
ightarrow-\infty}\frac{2+\frac{3}{x}}{1-\sqrt{4+\frac{3}{x^{2}}}}=\frac{2+0}{1 - 2}=- 2
$$

# Answer:
a) $-\frac{1}{2}$
b) $-\infty$
c) $0$
d) When $x
ightarrow\infty$, $\frac{2}{3}$; when $x
ightarrow-\infty$, $-2$

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QUESTION IMAGE

Question

Explanation:

Step1: Divide numerator and denominator by highest - power of \(x\)

Step2: Use \(\lim_{x

Step3: For part (b)

Step4: For part (c)

Step5: For part (d) (when \(x

Step6: For part (d) (when \(x

Answer: